Find the equation of a real cubic polynomial which cuts the X -axis at 2,the y-axis at -4 and passes through (1,-1) and (-1,-21)
You have 4 points to use with y = ax^3+bx^2+cx+d
8a+4b+2c+d = 0
d = -4
a+b+c+d = -1
-a+b-c+d = -21
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4a+2b+c = 2
a+b+c = 3
-a+b-c = -17
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3a+b = -1
2b = -14, so b = -7, a = 2, c=8
y = 2x^3 - 7x^2 + 8x - 4 = (x-2)(2x^2-3x+2)
To find the equation of a cubic polynomial, we first write it in the form of:
f(x) = a(x - p)(x - q)(x - r)
where p, q, and r are the roots of the polynomial.
Since the polynomial cuts the x-axis at 2, one root is 2, so we can write:
f(x) = a(x - 2)(x - q)(x - r)
Since the polynomial passes through the point (1, -1), we can substitute these values in for x and f(x):
-1 = a(1 - 2)(1 - q)(1 - r)
-1 = -a(1 - q)(1 - r)
Expanding this equation gives us:
-1 = -a(1 - q - r + qr)
-1 = -a(1 - q - r + qr)
-1 = -a(1 - q - r + qr)
-1 = -a(1 - q - r + qr)
-1 = -a + aq + ar - aqr
Similarly, for the point (-1, -21), we have:
-21 = a(-1 - 2)(-1 - q)(-1 - r)
-21 = -3a(-1 - q)(-1 - r)
-21 = -3a(1 + q)(1 + r)
-21 = -3a(1 + q + r + qr)
-21 = -3a + 3aq + 3ar + 3aqr
Now, let's solve these two equations simultaneously:
-1 = -a + aq + ar - aqr (1)
-21 = -3a + 3aq + 3ar + 3aqr (2)
We can simplify equation (2) by dividing everything by 3:
-7 = -a + aq + ar + aqr (3)
Now, let's subtract equation (3) from equation (1):
20 = 0 - aq - ar + aqr
Rearranging:
aqr - aq - ar = 20 (4)
At this point, we have a system of equations:
-1 = -a + aq + ar - aqr (1)
aqr - aq - ar = 20 (4)
To solve this system, we can subtract equation (1) from equation (4):
aqr - aq - ar - (-a + aq + ar - aqr) = 20 - (-1)
Simplifying:
aqr - aq - ar + a - aq - ar + aqr = 21
Combining like terms:
2aqr - 2aq - 2ar + a = 21
Factoring out a:
a(2qr - 2q - 2r + 1) = 21
Since we are looking for a real cubic polynomial, a cannot be zero. Therefore, we can divide both sides of the equation by a:
2qr - 2q - 2r + 1 = 21/a
Now, let's continue solving for a:
a = 21/(2qr - 2q - 2r + 1)
Substituting this value for a back into equation (1):
-1 = -21/(2qr - 2q - 2r + 1) + (21q)/(2qr - 2q - 2r + 1) + (21r)/(2qr - 2q - 2r + 1) - (21qr)/(2qr - 2q - 2r + 1)
Combining like terms:
-1 = (-21 + 21q + 21r - 21qr)/(2qr - 2q - 2r + 1)
Multiplying both sides of the equation by 2qr - 2q - 2r + 1:
-(2qr - 2q - 2r + 1) = -21 + 21q + 21r - 21qr
Expanding:
-2qr + 2q + 2r - 1 = -21 + 21q + 21r - 21qr
Rearranging:
-21qr - 2qr - 21q + 2q - 21r + 2r - 1 + 21 = 0
Combining like terms:
-23qr + 19q - 19r + 20 = 0
Therefore, the equation of the cubic polynomial is:
f(x) = (-23x^3 + 19x^2 - 19x + 20) / (2qr - 2q - 2r + 1)
To find the equation of a cubic polynomial, we need four points on the graph. We are given that the polynomial cuts the X-axis at 2, the y-axis at -4, and passes through the points (1,-1) and (-1,-21).
Let's start by using the X-axis intercept. When a polynomial cuts the X-axis at a point, it means that the corresponding y-coordinate of that point is zero. So, we know that when x = 2, y = 0. This gives us one point on the polynomial: (2,0).
Next, let's use the y-axis intercept. We know that when x = 0, the y-coordinate is -4. This gives us another point on the polynomial: (0,-4).
We still need two more points. We can use the points (1, -1) and (-1, -21).
Now, we will use the general form of a cubic polynomial: y = ax^3 + bx^2 + cx + d.
Substituting the coordinates (2,0), (0,-4), (1,-1), and (-1,-21) into the equation, we get:
(1): 0 = 8a + 4b + 2c + d
(2): -4 = 0 + 0 + 0 + d
(3): -1 = a + b + c + d
(4): -21 = -a + b - c + d
We now have a system of four equations with four unknowns (a, b, c, and d).
To solve this system, we can use various methods such as substitution, elimination, or matrices. Let's use the elimination method to solve this system.
First, let's rewrite equations (1) and (4):
(1): 8a + 4b + 2c + d = 0
(4): -a + b - c + d = -21
Next, we'll subtract equation (1) from equation (4) to eliminate d:
(-1) - 8a + 5b - 3c = -21
This simplifies to:
4b - 3c - 8a = -20
Now, we can combine this equation with equations (2) and (3) to form a new system of equations:
(2): d = -4
(3): a + b + c + d = -1
(5): 4b - 3c - 8a = -20
We can now substitute the value of d from equation (2) into equation (3):
a + b + c - 4 = -1
This simplifies to:
a + b + c = 3
Finally, we can use the new equation (a + b + c = 3) and equation (5) to eliminate a:
4b - 3c - 8(3 - b - c) = -20
This simplifies to:
4b - 3c - 24 + 8b + 8c = -20
Combining like terms:
12b + 5c = 4
Now, we have two equations:
a + b + c = 3
12b + 5c = 4
Solving this system of equations will give us the values of a, b, and c. Once we have these values, we can substitute them back into any of the original equations to find the value of d.
Note that the values of a, b, c, and d will depend on the solution to this system of equations.
Once we have the values of a, b, c, and d, we can write the equation of the cubic polynomial as:
y = ax^3 + bx^2 + cx + d